Question

x+2/x+3
Name the vertical asymptotes, if any
Name the horizontal asymptotes, if any

Answer 1

Vertical asymptotes are given by the domain. In your case, it would be -3 because you can't divide by 0. And horizontal asymptotes (also oblicuous) in this functions are given by the first coefficient, both top and down. In this case, it would be 1/1, which is 1, which means you have an horizontal asymptote on Y=1 , and a vertical on X=-3

Answer 2

What? No that's not true.

VA: x=-3
HA: y=0

because your HA is finding x for your denominator and y is your y-int. you do not have aa y-int. in this problem so y=0.

Answer 3

?? Dude, 0 isn't a HA. 1 is a HA because if it wasn't, then x+3 = x+2 , which is absurd...

Answer 4

NO. If it was x+2/x+3 + 1
then you would be right, but it's not.
x+2/x is your parent function. x+3 is how you get VA
you get HA by seeing whatever is added or subtracted to your x-int. which would be your y-int. y=0 because there's nothing added or subtracted in your y-int.

Answer 5

Please man, go read about homographic functions first. If not, go learn about something called limits, and see what happens with the function in infinite. There u have your definition of Horizontal Asymptote, and then come back and tell me that the horizontal asymptote is still 0..... i even explained you that x+3 != x+2, and therefore Y can never be 1 .....

Answer 6

Okay I'm in Honors Algebra 2; I just took a quiz on this Monday and made a 100 so don't you think I would know what I'm doing.

Answer 7

Haha, i'm just telling you that you are wrong. I even explained how Horizontal Asymptotes work. What does it matter if you are in honors Algebra 2 (wtf is that anyway?) or in highscool?

Answer 8

(x+2)/(x+3)

VA= when bottom = 0 so x+3=0 -> is -3
H= where your x is very very large. just pick a very large x and plug it into x. which is 1